document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=873126>Question 873126</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The perimeter of a football field is 340 yards. The length of the field is 20 years more than twice the width. What are the dimensions of the field?<BR>\n" );
document.write( "A. 100x70 yards<BR>\n" );
document.write( "B. 120x50 yards<BR>\n" );
document.write( "C. 200x140 yards<BR>\n" );
document.write( "D. 240 x 100 yards </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #526666 by </B><A HREF=/tutors/aboutme.mpl?userid=indra89811><B>indra89811(24)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=indra89811><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=indra89811><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=526666\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> How can length be 20 \"Years\" more? :P Haha, it'll be 20 yards...<BR>\n" );
document.write( "Let width=w<BR>\n" );
document.write( "Length=w+20<BR>\n" );
document.write( "2*(w+w+20)=340<BR>\n" );
document.write( "2(2w+20)=340<BR>\n" );
document.write( "4w+40=340<BR>\n" );
document.write( "4w=300<BR>\n" );
document.write( "w=75<BR>\n" );
document.write( "Length=75+20=95<BR>\n" );
document.write( "Options are incorrect. It will be length=95 yards and breadth=75 yards<BR>\n" );
document.write( "-------------------Peace out, hun :*------------------<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );